Percentage of the pea plants which would be homozygous recessive in `F_1` generation when tall `F_1` heterozygous pea plants are selfed will be:

Options

A. 0.25
B. 0.5
C. 1
D. 0.75

Explanation

To solve the problem of determining the percentage of homozygous recessive pea plants in the F1 generation when tall F1 heterozygous pea plants are selfed, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Genotypes**:
- In pea plants, the tall trait (T) is dominant over the short trait (t).
- The F1 generation from a cross of homozygous tall (TT) and homozygous short (tt) plants will all be heterozygous tall (Tt).

2. **Selfing the F1 Generation**:
- When we self the F1 heterozygous plants (Tt x Tt), we need to set up a Punnett square to determine the possible genotypes of the offspring.

3. **Set Up the Punnett Square**:
- The Punnett square will look like this:
```
T t
------------
T | TT | Tt |
------------
t | Tt | tt |
```

4. **Analyze the Results**:
- From the Punnett square, we can see the following genotypes:
- 1 TT (homozygous tall)
- 2 Tt (heterozygous tall)
- 1 tt (homozygous recessive)
- This gives us a total of 4 offspring.

5. **Calculate the Percentage of Homozygous Recessive Plants**:
- The homozygous recessive genotype (tt) appears 1 time out of 4 total genotypes.
- Therefore, the percentage of homozygous recessive plants is:
\[ \text{Percentage of tt} = \left( \frac{1}{4} \right) \times 100 = 25\% \]

### Final Answer:
The percentage of the pea plants which would be homozygous recessive (tt) in the F1 generation when tall F1 heterozygous pea plants are selfed is **25%**.

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